Practice question
Question
An aluminium wire of length 1.5m and cross-sectional area 2×10−6m2 is stretched by a force of 140N. If the Young's modulus of aluminium is 7×1010N/m2, what is the elongation?
Explanation
Young's modulus: Y = FLAΔL. Rearrange: ΔL = FLAY. Substitute: ΔL = 140×1.52×10−6×7×1010 = 2101.4×105 = 1.5×10−3m = 1.5mm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.5mm. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.